= Solution
The seminorms in $\mathcal R$ have inverse images of intervals that constrain one coordinate at a time. Finite intersections of these sets are exactly the standard basic neighbourhoods for the product topology, so $(X\times Y,\mathcal R)$ is the <product of locally convex spaces>. If $F\in(X\times Y)^*$, then
$$
f(x)=F(x,0),
\qquad
g(y)=F(0,y)
$$
belong to $X^*$ and $Y^*$ and $F(x,y)=f(x)+g(y)$. Conversely every such pair defines a continuous functional, so $(X\times Y)^*=X^*\oplus Y^*$.
For the open convex sets $K_1,\ldots,K_n$, consider
$$
C=\{(x_1-x_n,\ldots,x_{n-1}-x_n):x_i\in K_i\}\subseteq X^{n-1}.
$$
This set is open and convex, and $0\notin C$ because the $K_i$ have empty intersection. Separate $0$ from $C$ by a continuous linear functional on the product. By the dual description just proved, it has the form
$$
L(z_1,\ldots,z_{n-1})=\sum_{j=1}^{n-1}f_j(z_j)
$$
for $f_j\in X^*$, and $L$ is nonzero with one strict sign on $C$. Define
$$
T:X\to\mathbb R^{n-1},
\qquad
T(x)=(f_1(x),\ldots,f_{n-1}(x)).
$$
If some vector belonged to every $T(K_i)$, choose $x_i\in K_i$ with the same image. Then every $f_j(x_j-x_n)=0$, making $L(x_1-x_n,\ldots,x_{n-1}-x_n)=0$, a contradiction. Hence $\bigcap_iT(K_i)=\varnothing$, proving <finite-dimensional separation of open convex sets>.
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